mai em cần rồi, em cản ơn nhiều!

Nếu thực hiện chia theo lược đồ Hoocne thì kết quả như thế này:

\(f\left(x\right)=\left(x^2+2x-3\right)\left(x^3-2x^2+7x-23\right)+68\)

Hay \(f\left(x\right)\) chia \(x^2+2x-3\) được thương \(x^3-2x^2+7x-23\) và dư \(68\)

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

\(\left(3-x\right)\left(x+1\right)-\left(2.x\right)\left(x+2\right)-3\)

\(=3x+3-x^2-x-2x^2-4x-3=-3x^2-2x\)

\(a,=\left(-4\right)\\ b,=\left(-70\right)\\ c,2002+\left[45-5^2\right]=2002+\left[45-25\right]\\ =2002+20=2022\)

-4

-133

6026

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b) Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-1}{\sqrt{x}+1}\)

Thay x=3 vào B, ta được:

\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)

#include <bits/stdc++.h>

using namespace std;

long long n,i,s;

int main()

{

cin>>n;

if (n%2==0)

{

s=1;

for (i=1; i<=n; i++)

if (i%2==0) s=s*i;

cout<<s;

}

else 

{

s=1;

for (i=1; i<=n; i++)

if (i%2==1) s=s*i;

cout<<s;

}

return 0;

}

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)